A038504 -id:A038504 - cp's OEIS frontend

A038520 Number of elements of GF(2^n) with trace 1 and subtrace 0.

0, 1, 0, 3, 4, 6, 20, 28, 64, 136, 240, 528, 1024, 2016, 4160, 8128, 16384, 32896, 65280, 131328, 262144, 523776, 1049600, 2096128, 4194304, 8390656, 16773120, 33558528, 67108864, 134209536, 268451840, 536854528, 1073741824

Offset: 0

Frank Ruskey

Colin Barker, Table of n, a(n) for n = 0..1000
F. Ruskey, Number of irreducible polynomials over GF(2) with given trace and subtrace
F. Ruskey, Number of elements of GF(2^n) with given trace and subtrace
Index entries for linear recurrences with constant coefficients, signature (0,2,4).

Cf. A038504, A000749, A038518, A038519, A038521.

LinearRecurrence[{0,2,4},{0,1,0,3},40] (* Harvey P. Dale, Oct 15 2017 *)

concat(0, Vec(x*(1 + x^2) / ((1 - 2*x)*(1 + 2*x + 2*x^2)) + O(x^40))) \\ Colin Barker, Aug 02 2019

a(n) = C(n, r+0)+C(n, r+4)+C(n, r+8)+... where r = 1 if n odd, r = 3 if n even.
a(n) = 2*a(n-2) + 4*a(n-3), n > 3. - Paul Curtz, Feb 06 2008
From Colin Barker, Aug 02 2019: (Start)
G.f.: x*(1 + x^2) / ((1 - 2*x)*(1 + 2*x + 2*x^2)).
a(n) = (2^n + i*((-1-i)^n - (-1+i)^n)) / 4 for n>0, where i=sqrt(-1).
(End)

A091917 Coefficient array of polynomials (z-1)^n-1.

Original entry on oeis.org

1, -2, 1, 0, -2, 1, -2, 3, -3, 1, 0, -4, 6, -4, 1, -2, 5, -10, 10, -5, 1, 0, -6, 15, -20, 15, -6, 1, -2, 7, -21, 35, -35, 21, -7, 1, 0, -8, 28, -56, 70, -56, 28, -8, 1, -2, 9, -36, 84, -126, 126, -84, 36, -9, 1, 0, -10, 45, -120, 210, -252, 210, -120, 45, -10, 1, -2, 11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1

Offset: 0

Paul Barry, Feb 13 2004

The first element has been changed to 1 to produce an invertible matrix. Alternatively, this is the coefficient array for the polynomials P(z,n) = Product_{j=0..n-1} (z-(1+w(n)^j)) where w(n) = e^(2*Pi*i/n), i=sqrt(-1).
The row entries determine interesting recurrences. For instance, a(n) = 4a(n-1) + 6a(n-2) + 4a(n-3), a(0)=a(1)=a(2)=1, gives A038503. Sequences of the form a(n) = Sum_{k=0..n} (binomial(n,k) if k mod m = r, otherwise 0), for r=0..m-1, result. Equivalently, a(n) = Sum_{j=0..n-1} 2^n*(cos(Pi*j/m))^n*cos((n-2r)Pi*j/m)/m, r=0..m-1. These include A024493, A024494, A024495, A038503, A038504, A038505. The inverse matrix is A091918.
Triangle T(n,k), 0 <= k <= n, read by rows given by [ -2, 2, 1/2, -1/2, 0, 0, 0, 0, 0, ...] DELTA [1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 11 2007

			Rows begin:
  { 1},
  {-2,  1},
  { 0, -2,  1},
  {-2,  3, -3,  1},
  { 0, -4,  6, -4,  1},
  ...

Alois P. Heinz, Rows n = 0..140, flattened

T:= n-> `if`(n=0, 1, (p-> seq(coeff(p,z,i), i=0..n))((z-1)^n-1)):
seq(T(n), n=0..12);  # Alois P. Heinz, May 23 2015

Table[If[n == 0, 1, CoefficientList[(z-1)^n-1, z]], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 08 2016 *)

row(n) = if (n==0, 1, Vecrev((z-1)^n-1)); \\ Michel Marcus, May 23 2015

T(n,k) = T(n-1,k-1) + T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,1) = T(2,2) = 1, T(1,0) = T(2,1) = -2, T(2,0) = 0, T(n,k) = 0 for k > n or for k < 0. - Philippe Deléham, May 23 2015

G.f.: (1-2*x-x^2+x^2*y)/((x-1)*(-x+x*y-1)). - R. J. Mathar, Aug 11 2015

A094266 LQTL Lean Quaternary Temporal Logic: a terse form of temporal logic created by assigning four descriptors such that false, becoming true, true and becoming false are represented and become a linear sequence. In a branching tree two alternative are open, change or no change. The integer sequence above is the count of the row possibilities of the four states over successive iterations.

Original entry on oeis.org

1, 1, 0, 0, 1, 2, 1, 0, 1, 3, 3, 1, 2, 4, 6, 4, 6, 6, 10, 10, 16, 12, 16, 20, 36, 28, 28, 36, 72, 64, 56, 64, 136, 136, 120, 120, 256, 272, 256, 240, 496, 528, 528, 496, 992, 1024, 1056, 1024, 2016, 2016, 2080, 2080, 4096, 4032, 4096, 4160, 8256, 8128, 8128, 8256, 16512

Offset: 0

Robert H Barbour and L. D. Painter, Jun 01 2004

This is a table read by rows of length 4. Every row is formed from the previous one by the circular Pascal triangle-like rule: a, b, c, d -> d+a, a+b, b+c, c+d. Consider a labeled binary tree such that the root has label 0 and every node labeled k has children labeled k and (k+1) mod 4; the n-th row of this sequence counts nodes on the level n+1 with labels 0, 1, 2, 3, while the n-th row of A099423 counts nodes up to level n. - Andrey Zabolotskiy, Jan 06 2023

Robert H. Barbour, 2D Four-Color Cellular Automaton, NKS 2006 Wolfram Science Conference.
Robert H. Barbour, Two-dimensional Four Color Cellular Automaton: Surface Explorations, Complex Systems, 17 (2007), 103-112.

Cf. A000749, A005418, A038503, A038504, A038505, A020522, A063376.

Algorithm available from Robert H Barbour

Appears to satisfy a 12-degree linear recurrence. - Ralf Stephan, Dec 04 2004

A307078 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. ((1-x)^(k-2))/((1-x)^k-x^k).

Original entry on oeis.org

1, 1, 3, 1, 2, 7, 1, 2, 4, 15, 1, 2, 3, 8, 31, 1, 2, 3, 5, 16, 63, 1, 2, 3, 4, 10, 32, 127, 1, 2, 3, 4, 6, 21, 64, 255, 1, 2, 3, 4, 5, 12, 43, 128, 511, 1, 2, 3, 4, 5, 7, 28, 86, 256, 1023, 1, 2, 3, 4, 5, 6, 14, 64, 171, 512, 2047, 1, 2, 3, 4, 5, 6, 8, 36, 136, 341, 1024, 4095

Offset: 0

Seiichi Manyama, Mar 22 2019

			Square array begins:
     1,   1,   1,   1,  1,  1,  1,  1, 1, ...
     3,   2,   2,   2,  2,  2,  2,  2, 2, ...
     7,   4,   3,   3,  3,  3,  3,  3, 3, ...
    15,   8,   5,   4,  4,  4,  4,  4, 4, ...
    31,  16,  10,   6,  5,  5,  5,  5, 5, ...
    63,  32,  21,  12,  7,  6,  6,  6, 6, ...
   127,  64,  43,  28, 14,  8,  7,  7, 7, ...
   255, 128,  86,  64, 36, 16,  9,  8, 8, ...
   511, 256, 171, 136, 93, 45, 18, 10, 9, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns 1-6 give A126646, A000079, A024494(n+1), A038504(n+1), A133476(n+1), A119336.

Cf. A306846, A306915, A307079.

T[n_, k_] := Sum[Binomial[n+1, k*j+1], {j, 0, Floor[n/k]}]; Table[T[n-k, k], {n, 0, 12}, {k, n, 1, -1}] // Flatten (* Amiram Eldar, May 20 2021 *)

A(n,k) = Sum_{j=0..floor(n/k)} binomial(n+1,k*j+1).

A(n,2*k) = Sum_{i=0..n} Sum_{j=0..n-i} binomial(i,k*j) * binomial(n-i,k*j).

A099855 a(n) = n2^n - 2^(n/2)sin(Pi*n/4).

Original entry on oeis.org

0, 1, 6, 22, 64, 164, 392, 904, 2048, 4592, 10208, 22496, 49152, 106560, 229504, 491648, 1048576, 2227968, 4718080, 9960960, 20971520, 44041216, 92276736, 192940032, 402653184, 838856704, 1744822272, 3623870464, 7516192768

Offset: 0

Paul Barry, Oct 28 2004

Related to binomial transform of A002265. Sequence is identical to its fourth differences (cf. A139756, A137221). See also A097064, A135035, A038504, A135356. - Paul Curtz, Jun 18 2008

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-14,16,-8).

Cf. A001787, A002265, A009545, A038504, A097064, A135035, A135356.
Cf. A137221, A139756.
Binomial transform of A047538.

I:=[0,1,6,22]; [n le 4 select I[n] else 6*Self(n-1) -14*Self(n-2) +16*Self(n-3) -8*Self(n-4): n in [1..51]]; // G. C. Greubel, Apr 20 2023

LinearRecurrence[{6,-14,16,-8},{0,1,6,22},30] (* Harvey P. Dale, Mar 22 2018 *)

@CachedFunction
def a(n): # a = A099855
    if (n<5): return (0,1,6,22,64)[n]
    else: return 6*a(n-1) - 14*a(n-2) + 16*a(n-3) - 8*a(n-4)
[a(n) for n in range(51)] # G. C. Greubel, Apr 20 2023

G.f.: x/((1-2*x+2*x^2)*(1-4*x+4*x^2)).
a(n) = Sum_{k=0..n} 2^(k/2)*sin(Pi*k/4)*2^(n-k)*(n-k+1).
a(n) = 6*a(n-1) - 14*a(n-2) + 16*a(n-3) - 8*a(n-4).
a(n) = 2*A001787(n) - A009545(n).

A100216 Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).

Original entry on oeis.org

1, 4, 9, 16, 26, 44, 84, 176, 376, 784, 1584, 3136, 6176, 12224, 24384, 48896, 98176, 196864, 393984, 787456, 1573376, 3144704, 6288384, 12578816, 25163776, 50335744, 100675584, 201342976, 402661376, 805289984, 1610563584, 3221159936

Offset: 0

Creighton Dement, Nov 11 2004

A100215(n) (ves) = ((-1)^n)*A009116(n+3) (jes) + a(n) (les) + A038503(n+1) (tes) (Sn, below, corresponds to the generating function from above). Coefficients of Sn(z)*(1-z)/(1+z) gives match to A038504 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 1"). Coefficients of Sn(z)/(1+z) gives match to A038505 (Sum of every 4th entry of row n in Pascal's triangle, starting at "n choose 2"). Coefficients of Sn(z)/(1-z^2) gives match to A000749 (Number of strings over Z_2 of length n with trace 1 and subtrace 1). The elements 'i, 'j, 'k, i', j', k', 'ii', 'jj', 'kk', 'ij', 'ik', 'ji', 'jk', 'ki', 'kj', e ("floretions") are members of the quaternion product factor space Q x Q /{(1,1), (-1,-1)}. "les" sums over coefficients belonging to basis vectors which squared give the unit e (excluding e itself).

This sequence is identical to its 4th differences. - Jean-François Alcover, Nov 07 2013

			a(2) = 9 because (.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e)^3 =
1'j + 1'k + 1j' + 1k' + 3'ii' + 2'jj' + 2'kk' + 1'jk' + 1'kj' + 1e
and the sum of the coefficients belonging to basis vectors which squared give the unit e (excluding e itself) is 3+2+2+1+1 = 9 (see comment).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4).

Cf. A000749, A009116, A009545, A038503, A038504, A038505, A100213, A100214, A100215.

[n le 3 select n^2 else 4*Self(n-1) -6*Self(n-2) +4*Self(n-3): n in [1..40]]; // G. C. Greubel, Mar 28 2024

a:= n-> (<<0|1|0>, <0|0|1>, <4|-6|4>>^n. <<1, 4, 9>>)[1, 1]:
seq(a(n), n=0..35);  # Alois P. Heinz, Nov 07 2013

d = 4; nmax = 31; a[n_ /; n < d] := (n + 1)^2; seq = Table[a[n], {n, 0, nmax}]; seq /. Solve[ Thread[ Take[seq, nmax - d + 1] == Differences[seq, d]]] // First (* Jean-François Alcover, Nov 07 2013 *)
LinearRecurrence[{4,-6,4}, {1,4,9}, 41] (* G. C. Greubel, Mar 28 2024 *)

@CachedFunction
def a(n): # a = A100216
    if n<3: return (n+1)^2
    else: return 4*a(n-1) -6*a(n-2) +4*a(n-3)
[a(n) for n in range(41)] # G. C. Greubel, Mar 28 2024

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3), with a(0) = 1, a(1) = 4, a(2) = 9.
G.f.: (1-x^2)/((1-2*x)*(1-2*x+2*x^2)).
(a(n)) = lesseq(.5 'j + .5 'k + .5 j' + .5 k' + 1 'ii' + 1 e).
2*a(n) = 3*2^n - A009545(n+1) + 4*A009545(n). - R. J. Mathar, May 21 2019
E.g.f.: (1/2)*exp(x)*(3*sin(x) - cos(x) + 3*exp(x)). - G. C. Greubel, Mar 28 2024

A323346 Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = 1 (including e = 1) in Clifford algebra Cl(p,q)(R).

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 2, 6, 10, 10, 6, 6, 12, 16, 20, 16, 12, 16, 28, 28, 36, 36, 28, 28, 36, 64, 56, 64, 72, 64, 56, 64, 72, 136, 120, 120, 136, 136, 120, 120, 136, 136, 272, 256, 240, 256, 272, 256, 240, 256, 272, 256, 528, 528, 496, 496, 528, 528, 496, 496, 528, 528, 496

Offset: 0

Jianing Song, Jan 12 2019

See A323100 for an introduction of Clifford algebras.

			Table begins
p\q|  0   1   2    3    4    5  ...
---+-------------------------------
0  |  1,  1,  1,   2,   6,  16, ...
1  |  2,  3,  4,   6,  12,  28, ...
2  |  3,  6, 10,  16,  28,  56, ...
3  |  4, 10, 20,  36,  64, 120, ...
4  |  6, 16, 36,  72, 136, 256, ...
5  | 12, 28, 64, 136, 272, 528, ...
...
See A323100 for an example that shows T(1,3) = 6.

Jianing Song, Antidiagonals n = 0..99, flattened
Wikipedia, Clifford algebras

Cf. A038503(n+1) (first row), A038504(n+1) (first column), A007582 (main diagonal).

A323100 is the complement sequence.

s := sqrt(2): h := n -> [ 0, -s, -2, -s, 0, s, 2,  s][1 + modp(n+1, 8)]:
T := proc(n, k) option remember;
if n = 0 then return 2^k*(1 - 1/2) - 2^((k - 3)/2)*h(k + 2) fi;
if k = 0 then return 2^n*(1 - 1/2) - 2^((n - 3)/2)*h(n) fi;
T(n, k-1) + T(n-1, k) end:
for n from 0 to 9 do seq(T(n, k), k=0..9) od; # Peter Luschny, Jan 12 2019

T[n_, k_] := 2^(n + k) - Sum[Binomial[n, i] Binomial[k, j] Mod[Binomial[i - j, 2], 2], {i, 0, n}, {j, 0, k}];
Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] (* Jean-François Alcover, Jun 19 2019 *)

T(p,q) = sum(i=0, p, sum(j=0, q, binomial(p, i)*binomial(q, j)*!(binomial(i-j, 2)%2)))

T(p,q) = Sum_{i=0..p} Sum_{j=0..q} binomial(p, i)*binomial(q, j)*(1 - (binomial(i - j, 2) mod 2)).

T(p,q) = 2^(p+q) - A323100(p,q).

A130668 Diagonal of A129819.

Original entry on oeis.org

0, 0, 1, -2, 5, -11, 23, -48, 102, -220, 476, -1024, 2184, -4624, 9744, -20480, 42976, -90048, 188352, -393216, 819328, -1704192, 3539200, -7340032, 15203840, -31456256, 65010688, -134217728, 276826112, -570429440, 1174409216

Offset: 0

Paul Curtz, Jun 27 2007

sign

This sequence is connected to A124072. To see this, change the sign of every negative term and consider the differences of every line. Hence for the second line, and following lines, the four terms form periodic sequences:
 1   0   1   0
 0   0   1   1
 0   1   2   1
 1   3   3   1
 4   6   4   2
10  10   6   6
20  16  12  16
36  28  28  36
64  56  64  72
120 120 136 136
240 256 272 256.
The lines are connected as seen by the examples: (3rd line connected to 2nd, from right to left) 1+1=2, 1+0=1, 0+0=0, 0+1=1; (11th line connected to 10th) 136+136=272, 136+120=256, 120+120=240, 120+136=256.
The 4 columns are almost known (must the first line be suppressed?): A038503 (without the first 1), A000749 (without the first 0), A038505, A038504. Like the present sequence, every sequence of A124072 beginning with a negative number (-2, -11, ...) is a "twisted" sequence (see A129339 comments, A129961 and the present 4 columns). Periodic with period 2^n.
Inverse binomial transform of A129819. - R. J. Mathar, Feb 25 2009

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-6,-14,-16,-8).

a:=[-2,5,-11,23];; for n in [5..30] do a[n]:=-6*a[n-1]+-14*a[n-2] -16*a[n-3]-8*a[n-4]; od; Concatenation([0,0,1], a); # G. C. Greubel, Mar 24 2019

I:=[-2,5,-11,23]; [0,0,1] cat [n le 4 select I[n] else -6*Self(n-1) - 14*Self(n-2)-16*Self(n-3)-8*Self(n-4): n in [1..30]]; // G. C. Greubel, Mar 24 2019

gf = x^2*(1+x)*(1+3*x+4*x^2+3*x^3)/((1+2*x+2*x^2)*(1+2*x)^2); CoefficientList[Series[gf, {x, 0, 30}], x] (* Jean-François Alcover, Dec 16 2014, after R. J. Mathar *)
Join[{0, 0, 1}, LinearRecurrence[{-6,-14,-16,-8}, {-2,5,-11,23}, 30]] (* Jean-François Alcover, Feb 15 2016 *)

my(x='x+O('x^30)); concat([0,0], Vec(x^2*(1+x)*(1+3*x+4*x^2+3*x^3 )/((1+2*x +2*x^2)*(1+2*x)^2))) \\ G. C. Greubel, Mar 24 2019

(x^2*(1+x)*(1+3*x+4*x^2+3*x^3)/((1+2*x+2*x^2)*(1+2*x)^2 )).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 24 2019

From R. J. Mathar, Feb 25 2009: (Start)
G.f.: x^2*(1+x)*(1 + 3*x + 4*x^2 + 3*x^3)/((1 + 2*x + 2*x^2)*(1+2*x)^2).
a(n) = ((-1)^n*A001787(n+1) - 4*A108520(n) + 4*A122803(n))/32, n > 2. (End)
a(n) = -6*a(n-1) - 14*a(n-2) - 16*a(n-3) - 8*a(n-4) for n >= 7. - G. C. Greubel, Mar 24 2019

Extended by R. J. Mathar, Feb 25 2009

A137171 Interleaved reading of A000749 and its first to third differences.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 2, 1, 1, 3, 3, 1, 4, 6, 4, 2, 10, 10, 6, 6, 20, 16, 12, 16, 36, 28, 28, 36, 64, 56, 64, 72, 120, 120, 136, 136, 240, 256, 272, 256, 496, 528, 528, 496, 1024, 1056, 1024, 992, 2080, 2080, 2016, 2016, 4160, 4096, 4032, 4096, 8256, 8128, 8128

Offset: 0

Paul Curtz, May 11 2008

A000749 is identical to its fourth differences, which implies that the 2nd differences equal the 5th, the 3rd differences the 6th and so on and implies that each of the sequences of these differences obeys the recurrence a(n)=4a(n-1)-6a(n-2)+4a(n-3), n > 3.
The table containing A000749 and its first differences (essentially A038505), 2nd differences (A038504) and 3rd differences (A038503) as the 4 rows is
O, 0, 0, 1, 4, 10, 20, 36, 64, ...
0, 0, 1, 3, 6, 10, 16, 28, 56, ...
0, 1, 2, 3, 4, 6, 12, 28, 64, ...
1, 1, 1, 1, 2, 6, 16, 36, 72, ...
Columns sums are 1, 2, 4, 8, 16, 32 ... = 2^n =A000079. The sequence reads this table column by column.

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 4, 0, 0, 0, -6, 0, 0, 0, 4).

Cf. A137172, A137173.

Join[{0, 0, 0, 1},LinearRecurrence[{0, 0, 0, 4, 0, 0, 0, -6, 0, 0, 0, 4},{0, 0, 1, 1, 0, 1, 2, 1, 1, 3, 3, 1},59]] (* Ray Chandler, Sep 23 2015 *)

Edited by R. J. Mathar, Jun 28 2008

A130305 Triangle read by rows: A007318 * A130304 as infinite lower triangular matrices.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 6, 4, 1, 6, 10, 10, 5, 1, 12, 16, 20, 15, 6, 1, 28, 28, 36, 35, 21, 7, 1, 64, 56, 64, 71, 56, 28, 8, 1, 136, 120, 120, 135, 127, 84, 36, 9, 1, 272, 256, 240, 255, 262, 211, 120, 45, 10, 1

Offset: 1

Gary W. Adamson, May 20 2007

			First few rows of the triangle:
   1;
   2,  1;
   3,  3,  1;
   4,  6,  4,  1;
   6, 10, 10,  5,  1;
  12, 16, 20, 15,  6,  1;
  28, 28, 36, 35, 21,  7,  1;
  ...

Cf. A007318, A038504 (left column), A130304.

Binomial transform of A130304.

cp's OEIS Frontend

A038520 Number of elements of GF(2^n) with trace 1 and subtrace 0.

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A091917 Coefficient array of polynomials (z-1)^n-1.

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A307078 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. ((1-x)^(k-2))/((1-x)^k-x^k).

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A099855 a(n) = n*2^n - 2^(n/2)*sin(Pi*n/4).

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A100216 Relates row sums of Pascal's triangle to expansion of cos(x)/exp(x).

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A323346 Square array read by ascending antidiagonals: T(p,q) is the number of bases e such that e^2 = 1 (including e = 1) in Clifford algebra Cl(p,q)(R).

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Mathematica

PARI

A099855 a(n) = n2^n - 2^(n/2)sin(Pi*n/4).